2020 AMC 10B Problem 8

P1-10~4 minPrint

Triangles: Area & Pythagorean Circles Split into exhaustive cases and handle each

Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ , $Q$ , and $R$ is a right triangle with area $12$ square units?

Show solution without answering

Key insight

The area forces R to be exactly 3 units from line PQ. Then check each place the right angle could be: at P, at Q, or at R.

Solution

Use the area first. Take $PQ = 8$ as the base. Area $= \tfrac12 \times 8 \times \text{height} = 12$ , so the height is $3$ . So $R$ must be exactly $3$ units from line $PQ$ : on a line $3$ units above $PQ$ or on a line $3$ units below it.
Now ask where the right angle is. It could be at $P$ , at $Q$ , or at $R$ . Check each case.
Right angle at $P$ . Then $R$ lies on the line through $P$ that is perpendicular to $PQ$ . That line meets the "above" line once and the "below" line once. That gives $2$ points.
Right angle at $Q$ . The same reasoning gives $2$ more points.
Right angle at $R$ . A fact from geometry: the angle at $R$ is a right angle exactly when $R$ is on the circle that has $PQ$ as a diameter. That circle has its center at the midpoint of $PQ$ and radius $4$ .
Each of our two lines is $3$ units from the center, which is less than the radius $4$ . So each line cuts the circle in $2$ points, giving $4$ points in total.
No point was counted twice. The points from steps 3 and 4 are $4$ units along from the midpoint of $PQ$ and $3$ units up or down, so they are $\sqrt{4^2 + 3^2} = 5$ from the center. They are not on the circle of radius $4$ .

Total: $2 + 2 + 4 = 8$ locations.

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Why this works

Each condition sends $R$ to a set of possible places. The area sends it to two parallel lines. The right angle sends it to one of three places: a line through $P$ , a line through $Q$ , or the circle on $PQ$ . Counting where these meet, one case at a time, makes sure no position is missed. Always compare the distance ( $3$ ) with the radius ( $4$ ) to be sure the circle is really hit.

Alternative approach

Use coordinates. Put $P = (0, 0)$ and $Q = (8, 0)$ , so $R = (x, 3)$ or $(x, -3)$ .

Right angle at $P$ : $x = 0$ .
Right angle at $Q$ : $x = 8$ .
Right angle at $R$ : by the Pythagorean theorem, $RP^2 + RQ^2 = PQ^2$ , so $x^2 + 9 + (x - 8)^2 + 9 = 64$ . This simplifies to $x^2 - 8x + 9 = 0$ , so $x = 4 \pm \sqrt7$ .

That is $4$ values of $x$ , each with $2$ choices of $\pm 3$ , for $8$ points.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Take $PQ$ as the base. Area $12 = \tfrac12 \cdot 8 \cdot h$ gives $h = 3$ , so $R$ must lie on one of the two lines parallel to $PQ$ at distance $3$ from it.

Now decide where the right angle is.

Right angle at $P$ . Then $PR \perp PQ$ , so $R$ is on the perpendicular to $PQ$ at $P$ , at distance $3$ : two points (one on each side). Likewise right angle at $Q$ gives two more points. Subtotal: $4$ .

Right angle at $R$ . By Thales' theorem, $R$ lies on the circle with diameter $PQ$ , radius $4$ , centered at the midpoint of $PQ$ . Each parallel line at distance $3 < 4$ from the center crosses this circle in two points, so the two lines give $4$ points.

None of these coincide: the first four points are $\sqrt{4^2 + 3^2} = 5$ from the midpoint of $PQ$ , so they are not on the circle of radius $4$ . Total $4 + 4 = 8$ .

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 11: Circles
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)